1. The problem statement, all variables and given/known data An applied force of 30N acts on a 8kg box at an angle of 40 degrees above the horizontal. What is the acceleration of the block if =0.1 and μk = 0.05? What is the acceleration of the block without any friction? 2. Relevant equations Equation 1 F k = μ k F N Equation 2 F=ma Equation 3 3. The attempt at a solution Question 2 I attempted the second question first since it seemed to me easier. First I set up F=ma and plugged in cos(40)*30 since I need the x component of the applied force and then the mass which is 8 kg. Solving it gave me 2.9m/s2 which was the correct answer for the question. Question1 For the first question I began with identifying all the forces acting in the Y-direction since the normal force would be needed to calculate the static friction force with equation 1 : Fn , Fmg and the y component of the applied force Fa. Both Fmg and Fay act in the negative y direction so adding them up would equal Fn. Using trig functions I calculated Fay to be sin(40)30=19.28N and for Fmg I got 78.48N. Fn=78.48N + 19.28N= 97.76N Next I identified all forces acting in the x-direction which is the x-component of the applied force and the static frictional force acting in the opposite direction: Fax and Ffs. Then I used the sum of forces in the x- direction and got Fxsum=Fax + (-Ffs) = cos(40)*30N - (0.1*97.76N) = 22.98N - 9.776N = 13.204N So i think that's the remaining force after static friction has been overcome? Now I am a bit confused on how to continue. Do I now calculate the kinetic friction and the subtract it from 13.204N and then use equation 3 (F=ma) to solve for the acceleration? The given solutions were: acceleration without friction: 2.9m/s2 acceleration with friction : 2.3m/s2 This is my first post in the homework(first post ever in this forum) section and I apologize if I did not meet all of the Guidelines in this section (self-made graphic is too big?). I hope I didn't put too much stuff into this thread but I thought that if I put less, it would be harder for the forum members to understand what I am struggling with.